Question

The largest term of the sequence $\frac{1}{503},\frac{4}{524},\frac{9}{581},\frac{16}{692},\frac{25}{875}...$, is

• A. $\frac{36}{1148}$
• B. $\frac{48}{1509}$
• C. $\frac{49}{1529}$
• D. $\frac{64}{1509}$

Answer 1

Answer: (C)

$\frac{49}{1529}$

The given series is of the form $\frac{n^2}{500+3n^3}$
Let $f\left(x\right)=\frac{x^2}{500+3x^3}$
On differentiating w.r.t $x$,
$f^{\prime }\left(x\right)=\frac{2x(500+3x^3)-9x^4}{(500+3x^3{)}^2}$
To find the maximum value, let us find the critical points.
$f^{\prime }\left(x\right)=0$
$\Rightarrow 2x(500+3x^3)-9x^4$
$x={\left(\frac{1000}{3}\right)}^{\frac{1}{3}}$ or $x=0$
At $x={\left(\frac{1000}{3}\right)}^{\frac{1}{3}}$, $f^{\prime }$ changes value from negative to positive. Hence, it is a point of maxima.
We have,
$6<{\left(\frac{1000}{3}\right)}^{\frac{1}{3}}<7$
Let us consider the values of $f\left(6\right)$ and $f\left(7\right)$ to find the maximum value.
$f\left(7\right)=\frac{7^2}{500+3\times 7^3}=\frac{49}{1529}$
$f\left(6\right)=\frac{6^2}{500+3\times 6^3}=\frac{36}{1148}$
$f\left(7\right)>f\left(6\right)$
Hence, option C is correct